Problem: Multiply the following complex numbers: $({4-i}) \cdot ({-4+3i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4-i}) \cdot ({-4+3i}) = $ $ ({4} \cdot {-4}) + ({4} \cdot {3}i) + ({-1}i \cdot {-4}) + ({-1}i \cdot {3}i) $ Then simplify the terms: $ (-16) + (12i) + (4i) + (-3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -16 + (12 + 4)i - 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -16 + (12 + 4)i - (-3) $ The result is simplified: $ (-16 + 3) + (16i) = -13+16i $